دورات قصيرة الأمد

1. 1
Power Distribution SystemPower Factor ImprovementBYDr. MunthearAlqaderi

2. GENERAL EQUATION
FIG. 19.1 Defining the power delivered to a load.

3. RESISTIVE CIRCUIT
For a purely resistive circuit (such as that in Fig. 19.2), vand i are in phase, and θ= 0°, as appearing in Fig. 19.3.
FIG. 19.2 Determining the power delivered to a purely resistive load.

4. RESISTIVE CIRCUIT
FIG. 19.3 Power versus time for a purely resistive load.

5. RESISTIVE CIRCUIT
FIG. 19.4 Example 19.1.

6. RESISTIVE CIRCUIT
FIG. 19.5 Power curve for Example 19.1.

7. APPARENT POWER
FIG. 19.6 Defining the apparent power to a load.

8.  Most plant loads are Inductive and require a magnetic field to
operate:
 Motors
 Transformers
 Florescent lighting
 The magnetic field is necessary, but produces no useful work
 The utility must supply the power to produce the magnetic field and
the power to produce the useful work: You pay for all of it!
 These two types of current are the ACTIVE and REACTIVE
components
ACTIVE & REACTIVE POWERS

9. Advantages of Phasor Analysis
0
2 2
Resistor ( ) ( )
( )
Inductor ( )
1 1
Capacitor ( ) (0)
C
Z = Impedance
R = Resistance
X = Reactance
X
Z = =arctan( )
t
v t Ri t V RI
di t
v t L V j LI
dt
i t dt v V I
j C
R jX Z
R X
R




 
 
 
   


Device Time Analysis Phasor
(Note: Z is a
complex number but
not a phasor)

10. INDUCTIVE CIRCUIT AND REACTIVE POWER
FIG. 19.8 Defining the power level for a purely inductive load.

11. INDUCTIVE CIRCUIT AND REACTIVE POWER
FIG. 19.9 The power curve for a purely inductive load.  QSVARindP

12. CAPACITIVE CIRCUIT
For a purely capacitive circuit (such as that in Fig. 19.12), i leads vby 90°, as shown in Fig. 19.13.
FIG. 19.12 Defining the power level for a purely capacitive load.

13. CAPACITIVE CIRCUIT
FIG. 19.13 The power curve for a purely capacitive load. SQ  PVARcap

14. Power Factor Fundamental
Definitions:
Working /Active Power: Normally measured in kilowatts (kW). It does the "work" for the system-- providing the motion, torque, heat, or whatever else is required.
Reactive Power:Normally measured in kilovolt- amperes-reactive (kVAR), doesn't do useful "work." It simply sustains the electromagnetic field.
Apparent Power: Normally measured in kilovolt- amperes (kVA).Working Power and Reactive Power together make up apparent power.

15. Complex Power
 
*
cos( ) sin( )
P = Real Power (W, kW, MW)
Q = Reactive Power (var, kvar, Mvar)
S = Complex power (VA, kVA, MVA)
Power Factor (pf) = cos
If current leads voltage then pf is leading
If current
V I V I
V I
S V I j
P jQ
   

  




lags voltage then pf is lagging
(Note: S is a complex number but not a phasor)

16. Power Factor
P
V
P
I
I
V
(a) Purely Resistive Load (b) Resistive and Reactive Load

Pav = VavIav cos()

17. Power Factor:The Beer Analogy
Mug Capacity = Apparent Power (KVA)
Foam = Reactive Power (KVAR)
Beer = Real Power (kW)
Power Factor =
Beer (kW)
Mug Capacity (KVA)
Capacitorsprovide the Foam (KVAR), freeing up Mug Capacity so you don’t have to buy a bigger mug and/or so you can pay less for your beer !
kVAR
Reactive
Power
kW
Active
Power
kVA
Apparent
Power

18. 18
WHAT IS POWER FACTOR?
PowerFactoristheratioofACTIVEPOWERtotheTOTALPOWER(apparentpower):
=ActivePower=PTotalPowerS
S=TotalpowerofGenerator(orused)
P=Powerconsumedintheload(activepower)
Q=Reactive power stored in magnetic field. Or wasted power
Power Factor

19. 19
WHAT IS POWER FACTOR?
VectorialRepresentation:
S
P.Q
Φ
P
S
Q
j=90o
Generator
Load
Total power = S = VI= (units = KVA) Active power = P = VI CosΦ = (units = KW) Reactive power= Q = VI SinΦ = (units = KVAR)
Φ
V
I
V = Voltage : VoltsI = Current : AmpereΦ = Physical displacement of V&I
Power Factor = CosΦ

20. 20
WHAT IS LOW POWER FACTOR?
Iftheratioofactivepower(P)tototalpower(S)islessthanone(unity)thenthepowerfactorislow,whichmeanstotalpowerisnotbeingconsumed.
Generator
Generator
S = 100KVA
P = 80KW
P.F = 0.8
Q = 60-KVAR
S = 100KVA
P = 100KW
P.F = 1.0
Q = 0
Example:
P.F. =
P
S

21. 21
WHAT IS LOW POWER FACTOR?
Theaboveexampleclearlyindicatesthatageneratoroftotalpowerof100-KVAwillsupplymaximumof80-KWofactivepowertoaloadwithP.F.=0.8andthesamegeneratorwillsupplymaximumof100-KWofactivepowertoloadwithP.F=1.0.

22. 22
HOW TO IMPROVE THE POWER FACTOR ?
ThepowerfactorcanbeimprovedbysupplyingKVARtotheloads(inductivetype)
“CapacitorissourceofKVARs”
Thereforethepowerfactorofconnectedloadcanbeimprovedbyinstallingpowerfactorimprovementcapacitors/capacitorbanks

23. 59.7 kV
17.6 MW
28.8 MVR
40.0 kV
16.0 MW
16.0 MVR
17.6 MW 16.0 MW
28.8 MVR -16.0 MVR
Power System Notation
Power system components are usually shown as
“one-line diagrams.” Previous circuit redrawn
Arrows are
used to
show loads
Generators are
shown as circles
Transmission lines
are shown as a
single line

24. Reactive Compensation
44.94 kV
16.8 MW
6.4 MVR
40.0 kV
16.0 MW
16.0 MVR
16.8 MW 16.0 MW
6.4 MVR 0.0 MVR
16.0 MVR
Key idea of reactive compensation is to supply reactive
power locally. In the previous example this can
be done by adding a 16 Mvar capacitor at the load
Compensated circuit is identical to first example with
just real power load

25. Reactive Compensation, cont’d
Reactive compensation decreased the line flow from 564 Amps to 400 Amps. This has advantages
Lines losses, which are equal to I2 R decrease
Lower current allows utility to use small wires, or alternatively, supply more load over the same wires
Voltage drop on the line is less
Reactive compensation is used extensively by utilities
Capacitors can be used to “correct” a load’s power factor to an arbitrary value.

26. 26
HOW TO IMPROVE THE POWER FACTOR ?
LOAD
LOW POWER FACTOR
LOAD
IMPROVED POWER FACTOR
CAPACITOR
Fig.I

27. Calculation
If the original inductive load has apparent power S1, then
P = S1cos 1and Q1= S1sin 1= P tan 1
If we desired to increased the power factor from cos1to cos2without altering the real power, then the new reactive power is
Q2= P tan 2
The reduction in the reactive power is caused by the shunt capacitor is given by
QC= Q1–Q2= P (tan 1-tan 2)

28. 28
Calculation
rms
2
1 2
rms
2
C
ωV
P(tanθ tanθ )
ωV
Q
C

 
The value of the required shunt capacitance is
determined by the formula
Notice that the real power, P dissipated by the
load is not affected by the power factor correction
because the average power due to the capacitor
is zero

29. 29
Example 11.15
When connected to a 120V (rms), 60Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

30. 30
Solution
5000VA
0.8
4000
cos 1
1   

P
S
If the pf = 0.8 then,
cos1 = 0.8  1 = 36.87o
where 1 is the phase difference between the voltage and current.
We obtained the apparent power from the real power and the pf
as shown below.
The reactive power is
sin 5000sin 36.87 3000VAR 1 1 1 Q  S   

31. 31
Solution
4210.5VA
0.95
4000
cos 2
2   

P
S
When the pf raised to 0.95,
cos2 = 0.95  2 = 18.19o
The real power P has not changed. But the apparent power has
changed. The new value is
The new reactive power is
sin 1314.4VAR 2 2 2 Q  S  

32. 32
Solution
3000 1314.4 1685.6VAR 1 2 Q Q Q    C
The difference between the new and the old reactive power is
due to the parallel addition of the capacitor to the load.
The reactive power due to the capacitor is
The value of capacitance
added is
310.5μF
2 (60)(120)
1685.6
2 2
  
 rms 
C
V
Q
C

33. In this example, demand was reduced to 8250 kVA from 10000 kVA. 1750KVA Transformer Capacity Release. The power factor was improved from 80% to 97%
Before
After
Why do we install Capacitors?

34. 34
DISADVATAGES OF LOW POWER FACTOR
1.Foragivenpowertobesupplied,thecurrentisincreased.
2.Thecurrentthusincreasedin-returncausesincreaseincopperlosses(PL=I2R)anddecreaseintheefficiencyofbothapparatusandthesupplysystem,whichresultsinoverloadingandhenceburningoftheassociatedequipment.

35. 35
DISADVATAGES OF LOW POWER FACTOR
3.Copperlossesintransformersalsoincreases.
4.Generators,transformers,switches,transmissionlinesandotherassociatedswitchgearbecomesover-loaded.
5.Voltageregulationofgenerators,transformersandtransmissionlinesincreases.
6.Hence,costofgeneration,transmissionanddistributionincreases.

36. 36
NATURAL POWER FACTORS
oCEILING FAN 0.5 TO 0.7
oCABIN FAN0.5 TO 0.6
oEXAUST FAN0.6 TO 0.7
oSEWING MACHINE0.6 TO 0.7
oWASHING MACHINE0.6 TO 0.7
oRADIO0.9
oVACUUM CLEANER0.6 TO 0.7
oTUBE LIGHT0.5 TO 0.9
oCLOCK0.9
oELECTRONIC EQUIPMENT0.4 TO 0.95

37. 37
NATURAL POWER FACTORS
oNEON SIGN0.5 TO 0.55
oWINDOW TYPE AIR CONDITIONER0.62 TO 0.85
oHAIR DRYERS0.7 TO 0.8
oLIQUIDISER0.8
oMIXER0.8
oCOFFEE GRINDER0.75
oREFRIGERATOR0.65
oFREEZER0.7
oSHAVER0.6
oTABLE FAN0.5 TO 0.6

38. 38
NATURAL POWER FACTORS
oMERCURY VAPOUR LAMPO.4 TO 0.6
oINDUSTRIAL INDUCTION MOTOR:
◘NO LOADO.18
◘25% FULL LOAD0.56
◘75% FULL LOAD0.81
◘100% FULL LOAD0.85
◘125% FULL LOAD0.86
oCOLD STORAGEO.76 TO 0.80
oCINEMAS0.78 TO 0.80
oMETAL PRESSINGO.57 TO 0.72

39. MOTOR LOAD CHARACTERISTICS

40. 40
NATURAL POWER FACTORS
oOIL MILLSO.51 TO 0.59
oWOOLEN MILLSO.70
oPOTTERIES0.61
oCIGARETTE MANUFACTURING 0.80
oFOUNDRIES0.59
oSTRUCTURAL ENGINEERING0.53 TO 0.68
oCHEMICALS0.72 TO 0.87
oMUNICIPAL PUMPING STATIONS0.65 TO 0.75
oOIL TERMINALS0.64 TO 0.83
oROLLING MILLS0.60 TO 0.72

41. 41
NATURAL POWER FACTORS
oPLASTIC MOLDING0.57 TO 0.73
oFILM STUDIOSO.65 TO 0.74
oHEAVY ENGINEERING WORK0.48 TO 0.75
oRUBBER EXTRUSION AND MOLDING0.48
oPHARMACEUTICALS0.75 TO 0.86
oOIL AND PAINT MANUFACTURING 0.51 TO 0.69
oBISCUIT FACTORY0.60
oLAUNDRIES0.92
oFLOUR MILLS 0.61
oGLASS WORKS 0.87

42. 42
NATURAL POWER FACTORS
oIRRIGATIONS PUMPSO.62 TO 0.80
oREPAIR SHOP, AUTOMATIC LATHE,0.6
WORKSHOP, SPINNING MILLS,
WEAVING MILL
oWELDING SHOP0.5 TO 0.6
oHEAT TREATMENT SHOP, STEEL0.65 TO 0.8
WORKS, ROLLING MILLS
oTEXTILE0.65 TO 0.75
oCEMENT0.8 TO 0.85
oOFFICE BUILDING O.8 TO 0.85

43. Three Options for Applying Power Factor Capacitors:
A) Fixed capacitors @ individual motors or @ MCC
B) Automatic Banks at Main Switch Board
C) De-tuned Automatic Capacitor Bank at Main Switch Board
M
M
M
M
M
A
B
C
A
Harmonic Source e.g. Variable Speed Drive
Capacitor Locations

44. 44
ADVANTAGES OF POWER FACTOR IMPROVEMENT
i.Addingcapacitor,releasescircuitcapacityformoreloadorrelievestheoverloadedcircuit.ThecapacitorKVARperKVAofloadincreaseisofparticularinterestasthisestablishestheaveragecostofsupplyingeachadditionalKVAofload. ThiscostcanbecomparedwiththecostperKVAofincreasingthetransformerorsupplycircuitratingandwouldjustifytheapplicationofcapacitors.
PFI Capacitor’s addition, thus can be viewed in two lights.

45. 45
ADVANTAGES OF POWER FACTOR IMPROVEMENT
ii.CapacitorsappliedtogivenloadreducetheI2Rlossesinthesupplycircuit.Fora70percentpowerfactorloadwith40-KVARofcapacitorsaddedforeach100KVAofcircuitcapacity,theI2Rlosswillbe59%ofitsformervalue.Thelossesarenotonlyreducedinthecircuitinwhichthecapacitorsareappliedbutinallthecircuitbacktoandincludingthesourcegenerator.

46. 46
ADVANTAGES OF POWER FACTOR IMPROVEMENT
AutomaticPowerFactorimprovementcapacitorsorcapacitorbanksappliedontheloadendofcircuit, withlaggingpowerfactor(morethan95%loads), haveparticulareffects,oneormoreofwhichmaybethereasonfortheapplication.
1.Improves the power factor at the source.
2.Reduces system losses as current in conductors decreases.

47. 47
ADVANTAGES OF POWER FACTOR IMPROVEMENT
3.Improves voltage level at the load.
4.Decreases KVA loading on the source.
5.Reduces investment in system facilities per KW of load supplied.